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rowdycar313p0ao5k
30.11.2019 •
Mathematics
Afood snack manufacturer samples 15 bags of pretzels off the assembly line and weighed their contents. if the sample mean is 10.2 and the sample standard deviation is 0.25, find the 95% confidence interval of the true
a. (10.06, 10.34)b. (10.07, 10.33)d. (10.14, 10.26)
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Ответ:
the correct option is A
Step-by-step explanation:
We want to find 95% confidence interval for the mean of the weight of pretzels.
Number of samples. n = 15 bags
Mean, u = 10.2
Standard deviation, s = 0.25
We will use the t- test
Degree of freedom = n - 1 = 15 - 1= 14
Alpha, a = (1-confidence interval )/2
a = (1-0.95)/2 = 0.025
Looking at the t-distribution table, the corresponding z value is 2.131
Confidence interval = z × standard deviation/√n
Confidence interval = 2.131 × 0.25/√15
Confidence interval = 0.13755545851
Approximately 0.138
At 95% confidence interval,
The lower end is 10.2 - 0.138 = 10.062 Approximately 10.06
The upper end is 10.2 - 0.138 = 10.338. Approximately 10.34
Ответ:
x – 4y = 8
Step-by-step explanation:
I just took the quiz :) and got 1000000000000000000000000%